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四川省绵阳市高中2022级第二次诊断性考试数学答案_2025年1月_250117四川省绵阳市高中2022级第二次诊断性考试(全科)_四川省绵阳市高中2022级第二次诊断性考试数学

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四川省绵阳市高中2022级第二次诊断性考试数学答案_2025年1月_250117四川省绵阳市高中2022级第二次诊断性考试(全科)_四川省绵阳市高中2022级第二次诊断性考试数学
四川省绵阳市高中2022级第二次诊断性考试数学答案_2025年1月_250117四川省绵阳市高中2022级第二次诊断性考试(全科)_四川省绵阳市高中2022级第二次诊断性考试数学
四川省绵阳市高中2022级第二次诊断性考试数学答案_2025年1月_250117四川省绵阳市高中2022级第二次诊断性考试(全科)_四川省绵阳市高中2022级第二次诊断性考试数学
四川省绵阳市高中2022级第二次诊断性考试数学答案_2025年1月_250117四川省绵阳市高中2022级第二次诊断性考试(全科)_四川省绵阳市高中2022级第二次诊断性考试数学
四川省绵阳市高中2022级第二次诊断性考试数学答案_2025年1月_250117四川省绵阳市高中2022级第二次诊断性考试(全科)_四川省绵阳市高中2022级第二次诊断性考试数学
四川省绵阳市高中2022级第二次诊断性考试数学答案_2025年1月_250117四川省绵阳市高中2022级第二次诊断性考试(全科)_四川省绵阳市高中2022级第二次诊断性考试数学
四川省绵阳市高中2022级第二次诊断性考试数学答案_2025年1月_250117四川省绵阳市高中2022级第二次诊断性考试(全科)_四川省绵阳市高中2022级第二次诊断性考试数学

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高中 2022 级第二次诊断性考试 数学参考答案及评分标准 一、选择题:本题共8小题,每小题5分,共40分. 1.D 2.A 3.C 4.B 5.D 6.B 7.A 8.C 二、选择题:本大题共3小题,每小题6分,共18分.在每小题给出的四个选项中,有多项符合题目 要求.全部选对的得6分,选对但不全的得部分分,有选错的得0分. 9.BCD 10.BC 11.ABD 三、填空题:本题共3个小题,每小题5分,共15分. 12.3; 13. ; 14. 四、解答题:本题共5小题,第15题13分,第16、17小题15分,第18、19小题17分,共77分.解 答应写出文字说明、证明过程或演算步骤. 5.解:(1)∵ , 1 由正弦定理得: ,····························································2分 又 则 , , ∴ ,····················································································4分 ∴ ,又B是三角形内角,··········································································5分 ∴ ;········································································································6分 2)∵ ,且 , ( ∴ ,···································································································8分 ∴ ,·························································································9分 ∴ ,·························································································11分 ∴ .····································································································13分 16.解:(1)a=0时, , ,且 ,······························2分 ∴ ,·································································································4分 故切线方程为:y−(e+1)=e(x−1),即ex−y+1=0;·························································6分 (2)∵ , ,·········································································7分 由10),易知 , , 设平面ACE的法向量为:n=(x,y,z), 1 ∴ ,不妨令 ,则平面ACE的一个法向量为:n=(0,−n,m), 1 又平面ACD的一个法向量为n=(0,0,1),·································································7分 2 ∴cos= ,······························································8分 1 2 ∵ ,则 , 解得:n= ,则点E到平面ACD的距离为 , 由E为PD的中点,则点P到平面ACD的距离为 ,·················································9分 在△ACD中,AC=AD=2,满足 , ∴AD⊥AC,则△ACD的面积为2,···········································································10分 ∴三棱锥P-ACD的体积 ;··························································11分 (3)由PC//EG,则直线PC与平面ACE所成角即为直线EG与平面ACE所成角,设为θ, 由 ,则 ,·································································12分 ·····················13分 ∴ ······························14分 ∴ (当且仅当m=1时,等号成立)·······················16分 数学参考答案 第5页(共6页)即 的最大值为 , ∴直线PC与平面ACE所成角正弦值的最大值为 .················································17分 19.解:(1)由已知得, ,即 ,······················································1分 又离心率为 ,则 , ∵ ,所以 ,即 ,·····························································2分 ∴ , ,·································································································3分 ∴椭圆Г的标准方程为: ;······································································4分 (2)设点M(x,y),则点M满足: ,则 , 0 0 由已知可得F(0, ),H(0, ),设直线MF与MH的斜率分别为 , , ∴ , ,···············································································5分 直线MF与MH的斜率之积满足: .······················6分 (i)∵D(2, ),G(2,0),则 , , 直线PF的方程为: ,令 ,则 , ∴ ,·····································································································7分 直线HQ的方程为: ,令 ,可得Q(2, ), ∴ , ∴ ,················································································8分 且 ,·························································9分 ∴ ;···················································································10分 (ii) 存在 ,使得|TK|定值 ,理由如下:···················································11分 设点 , , , ①当过椭圆上点 的直线l斜率存在时,设直线l方程为: , 数学参考答案 第6页(共6页)带入椭圆Г的方程: , 化简并整理得: , ∵直线l与椭圆仅有一个公共点, ∴ , 化简得: ,····················································································12分 ∴ ,代入 ,得 , ∴ , 从而直线l的方程为: ,即 ,·············································13分 ②当过 的直线l斜率不存在且与椭圆Г仅有一个公共点时,直线l的方程为: 满足上式. 同理:当过椭圆上点 的直线: 与椭圆Г仅有一个公共点, 这两条直线都过点S,所以有 , , ∴直线IJ的方程为 .···········································································14分 由(i)则直线PF的方程为: ,令 ,则R( , ), 又Q(2, ), ∴RQ的中点S( , ),即 , . ∴直线IJ的方程表示为: , 即 ,·······································································15分 令 ,解得: ,·······························································16分 ∴直线IJ恒过定点N( ),又∵OT⊥IJ, ∴点T在以ON为直径的圆上,即K( ),|TK|定值 .·······································17分 数学参考答案 第7页(共6页)